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 #2
avatar+1616 
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Here's how to find the equation for the distance an object falls before reaching terminal velocity and estimate the distance for a piece of notebook paper

 

1. Combining Equations:

 

We are given two equations

 

Speed due to gravity: |v| = gt (where v is the object's speed, g is acceleration due to gravity, and t is time)

 

Terminal velocity: v_terminal = √(2mg / ρA) (where m is the object's mass, ρ is air density, A is the object's cross-sectional area, and C_D is the drag coefficient, assumed to be ≈ 1)

 

We want to find the time (t) it takes for the object's speed to reach terminal velocity (v_terminal).

 

2. Solving for Time:

 

Since the object accelerates constantly until reaching terminal velocity, we can set its speed (v) equal to the terminal velocity (v_terminal) in the first equation:

 

v_terminal = gt

 

Substitute the expression for terminal velocity from the second equation:

 

√(2mg / ρA) = gt

 

3. Square Both Sides (be cautious):

 

Square both sides to get rid of the square root (remembering that squaring introduces extraneous solutions, so we'll need to check for those later):

 

2mg / ρA = g^2 * t^2

 

4. Isolate Time:

 

Solve for time (t):

 

t = √(2mg / (ρA * g^2))

 

5. Estimating Distance:

 

Once we have the time (t), we can estimate the distance (d) the object falls by multiplying the terminal velocity (v_terminal) by the time (t):

 

d = v_terminal * t = √(2mg / ρA) * √(2mg / (ρA * g^2))

 

Simplify the equation:

 

d = √((2mg)^2 / (ρA * g^2 * ρA))

 

Cancel out common factors and remove the square root (since distance cannot be negative):

 

d = 2m / (ρA)

 

6. Estimating for Notebook Paper (as an example):

 

Let's estimate the distance for a piece of notebook paper (assuming no wind resistance and the chosen estimates for mass, density, and area):

 

Mass of a sheet of notebook paper (m): Assume m ≈ 0.005 kg (This is an estimate, the actual mass can vary)

 

Air density (ρ): ρ ≈ 1.2 kg/m³ (at sea level)

 

Area of a sheet of notebook paper (A): Assume a typical notebook paper size of 21.6 cm x 27.9 cm. Convert to meters: A ≈ 0.06 m x 0.08 m = 0.0048 m²

 

Plug these values into the equation:

 

d = 2 * 0.005 kg / (1.2 kg/m³ * 0.0048 m²)

 

d ≈ 0.83 meters

17.04.2024
 #1
avatar+1616 
0

Here's how to find the length of PA (the distance between point P and point A) in the given scenario:

 

Properties of an Equilateral Triangle:

 

Since ABC is an equilateral triangle, all three sides (AB, AC, and BC) are equal in length.

 

Triangle Angle Bisection by Altitude:

 

The altitude drawn from a vertex of an equilateral triangle bisects the opposite side and also creates two 30-60-90 right triangles.

 

Applying Triangle Properties:

 

Let D be the midpoint of BC. Since the altitude from A bisects BC, point D coincides with the midpoint of segment PA.

 

We are given that PB = 18 and PC = 7. Since BC is divided into two segments with a ratio of 18:7, segment BD must have a length of 18 and segment CD a length of 7.

 

30-60-90 Right Triangle

:

Triangle BDP is a 30-60-90 right triangle because BD is half the hypotenuse of the equilateral triangle (BC), and the altitude from A bisects the base at a 60-degree angle (property of equilateral triangles).

 

In a 30-60-90 triangle, the shorter leg (opposite the 30-degree angle) is half the length of the hypotenuse. In this case, BD (shorter leg) = 18, so the hypotenuse (BP) is twice that length, which is 36.

 

Finding Segment AD:

 

Since triangle BDP is 30-60-90, the longer leg (opposite the 60-degree angle) is equal to the shorter leg multiplied by the square root of 3. We know the shorter leg (BD) is 18, so:

 

AD (longer leg) = BD * √3 = 18 * √3

 

Finding Segment PA:

 

Since D is the midpoint of segment PA, then PD = DA = (18 * √3) / 2

 

Now we can find the total length of PA by adding the lengths of segments PD and DA:

 

PA = PD + DA = (18 * √3) / 2 + (18 * √3) / 2

 

PA = 18√3 (We can simplify this further if needed, but the answer accepts sqrt(3))

 

Answer:

 

The length of PA is 18√3.

17.04.2024
 #1
avatar+1616 
0

Absolutely, I’ve been improving my problem-solving abilities in solving polynomial equations. Let's find a3+b3, where a and b are the roots of the equation: 5x2−11x+4=−3x2−17x+5

We can solve the equation for a and b using the quadratic formula and then use the fact that for any quadratic equation ax2+bx+c=0, the sum of the roots is −ab​ and the product of the roots is ac​.

Steps to solve: 1. Solve the equation for a and b: 5x2−11x+4=−3x2−17x+5

Combining like terms and rearranging the equation, we get: 8x2+6x−1=0

Using the quadratic formula, we get: x=2a−b±b2−4ac​​ where a, b, and c are the coefficients of the quadratic equation. In this case, a = 8, b = 6, and c = -1. Substituting these values into the formula, we get:

x=2⋅8−6±62−4⋅8⋅−1​​

x=16−6±217​​

Therefore, the roots are x=817​−3​ and x=8−17​−3​.

2. Find a^3 + b^3: We know that for any quadratic equation ax2+bx+c=0, the sum of the roots is −ab​ and the product of the roots is ac​. In this case, we have:

a+b=−86​=−43​ ab=8−1​

We can use these relationships to find a3+b3:

a3+b3=(a+b)(a2−ab+b2)

Substitute the values we found for a + b and ab:

a3+b3=−43​(a2−ab+b2)

We don't need to find the individual values of a^2 and b^2$ since we can rewrite the expression using the fact (a+b)2=a2+2ab+b2 :

(a+b)2=a2+2ab+b2

Substitute a + b = -3/4:

(−43​)2=a2+2ab+b2

Expand:

169​=a2+2ab+b2

Substitute ab = -1/8:

169​=a2−41​+b2

Combine like terms:

169​+41​=a2+b2

a2+b2=45​

Substitute this back into the expression for a^3 + b^3:

a3+b3=−43​(45​−81​)

a3+b3=−43​⋅89​

a3+b3=−3227​

Answer: a3+b3=−3227​

02.04.2024